Let $(G, *)$ be a group and $(H, *)$ its normal subgroup. Then the factor group is defined as
$$ G/H := \left\{g*H: g\in G \right\} .$$
We define the multiplication of cosets (elements of G/H) as a function $$ f: G/H \times G/H \rightarrow G/H $$given by $ \left(g_1 H, g_2 H \right) \mapsto (g_1 * g_2)*H $. Usually, this is shown to be well-defined function by using the "definition":
A map $f: A\rightarrow B$ is well-defined if for every $a, b \in A$ $a=b$ implies $f(a)=f(b)$.
However, this "definition" does not make sense to me: If $a=b$, doesn't $ f(a)=f(b)$ always follow, since I can always rewrite $a$ as $b$?
That is why I am hoping to find a more "explicitly" defined function which defines the same multiplication of cosets. To do this, I was trying to find a function $\tilde{f}: G/H \rightarrow G$ which maps every $g*H \in G/H$ satisfying $\hat{g}*H=g*H $ to $\hat{g}$. After that, I could define another function $ \overline{f}: G \rightarrow G/H $ given by $g \mapsto g*H$. Then I redefine the function $f$ as $ (x, y) \mapsto \overline{f}(\tilde{f}(x)*\tilde{f}(y)) $, which is clearly well-defined.
Now the problem is, can such function $\tilde{f} $ be found?
Best Answer
$G$ is divided into disjoint subsets $gH$. You can pick one element $\tilde g$ in every coset $gH$ (which can be done using AC). Then define $\tilde f(gH)$ as $\tilde g$. This can be done in many ways unless $H=\{1\}$.